U.U.D.M. Project Report 2016:45

Examensarbete i matematik, 15 hpHandledare: Jörgen ÖstenssonExaminator: Veronica Crispin QuinonezDecember 2016

Department of MathematicsUppsala University

Asymptotic Expansions of Integrals and theMethod of Steepest Descent

Erik Falck

Asymptotic Expansions of Integrals and the

Method of Steepest Descent

Erik Falck

December 16, 2016

Abstract

This paper gives an introduction to some of the most well-known meth-ods used for finding the asymptotic expansion of integrals. We start bydefining asymptotic sequences and asymptotic expansion. The classicalresult Watson’s lemma is discussed and a proof of Laplace’s method ispresented. The theory of the method of steepest descent, one of the mostwidely used techniques in asymptotic analysis is studied. This methodis then applied to calculate the asymptotics of the Airy function and thelinearized KdV equation.

1

Contents

1 Introduction 3

2 Elementary asymptotics and Watson’s lemma 42.1 Asymptotic expansion . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Watson’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3 Laplace’s Method 10

4 Method of Steepest Descent 134.1 The idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2 Steepest descent paths and saddle points . . . . . . . . . . . . . . 134.3 Leading order behavior of F (λ) . . . . . . . . . . . . . . . . . . . 16

5 Applications 185.1 Airy functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

5.1.1 Asymptotics as x −→ +∞ . . . . . . . . . . . . . . . . . . 205.1.2 Asymptotics as x −→ −∞ . . . . . . . . . . . . . . . . . . 24

5.2 The KdV equation . . . . . . . . . . . . . . . . . . . . . . . . . . 265.3 Concluding remarks . . . . . . . . . . . . . . . . . . . . . . . . . 30

6 Acknowledgments 30

7 References 31

2

1 Introduction

Many problems in mathematics and physics have solution formulas that can berepresented as integrals depending on a parameter, and one is often interestedin the behavior of the solution as the parameter λ goes to infinity. The mostfrequently used way of evaluating how these integral representations behave isto find an asymptotic expansion in the parameter λ. In many problems thesesolutions can be written as an integral of the general form

F (λ) =

∫C

eλR(z)g(z)dz (1.1)

where C is a contour in the complex plane. Moreover, many integral transformssuch as the Laplace transform and the Fourier transform have this form. Thispaper will give a brief introduction to some of the most common methods forfinding the asymptotic expansion of such integral representations, with focus onthe method of steepest descent and some applications of it.For more detailed information about this topic the text [8] by Peter Miller isespecially recommended. See also the following well-known texts: [6] by CarlM. Bender and Steven A. Orszag, [5] by Norman Bleistein and Richard A. Han-delsman, [1] by Mark J. Ablowitz and A. S. Fokas.

In Section 2 the definition of asymptotic expansion and some notation will bebrought up before going into the part of the methods where the classical resultknown as Watson’s lemma will be presented. In Section 3 we explain Laplace’smethod, a technique for finding the (dominant) contribution of the asymptoticsof a real valued integral coming from a point or points on the interval whereR(t) attains its maximum.In Section 4 the method of steepest descent is discussed, which can be consid-ered a generalization of Laplace’s method used for complex integrals. In Section5 two applications are presented, the asymptotics of the Airy function and theasymptotics of the linearized KdV equation, both computed using the methodof steepest descent.

3

2 Elementary asymptotics andWatson’s

lemma

2.1 Asymptotic expansion

To begin with there are a couple of definitions and expressions that needs to bebrought up to clarify the material in upcoming sections. The asymptotic be-havior of a function is expressed as an asymptotic expansion given a sequence offunctions, so to be able to form an asymptotic expansion one needs asymptoticsequences.

For the following definitions let D be a subset of the complex plane.

Definition 2.1. (Asymptotic sequence.) A sequence of functions{φn(z)

}∞n=0

is called an asymptotic sequence as z −→ z0 from D if whenever n > m, wehave φn(z) = o(φm(z)) as z −→ z0 from D.

Example 2.2. The sequence of functions{zn}∞n=0

is an asymptotic sequence

as z −→ 0, and{z−n

}∞n=0

is an asymptotic sequence as z −→∞.

Definition 2.3. (Asymptotic expansion.) Let{φn(z)

}∞n=0

be an asymptoticsequence as z −→ z0 from D. Then the sum

N∑n=0

anφn(z)

is said to be an asymptotic approximation as z −→ z0 from D of a function f(z)if

f(z)−N∑n=0

anφn(z) = o(φN (z))

as z −→ z0 from D. If{an}∞n=0

is a sequence of complex constants such thatthe above is true for each N, then the formal infinite series

∞∑n=0

anφn(z)

is called an asymptotic series and is said to be an asymptotic expansion of f(z)as z −→ z0 from D.

Normally the asymptotic expansion of f(z) is written in the following way:

f(z) ∼∞∑n=0

anφn(z)

4

as z −→ z0.Note that the symbol ∼ is used here. That is because the series does not nec-essarily need to converge and the use of an equality sign would therefore beincorrect. Even if the the series is divergent it will still say a lot about thebehavior of the function f(z).

Example 2.4. Consider a function f being C∞ in a neighborhood of the origin.Then from Taylor’s theorem we have the asymptotic expansion

f(x) ∼∞∑n=0

f (n)(0)

n!xn

as x −→ 0. Indeed,

f(x)−N∑n=0

f (n)(0)

n!xn = En(x),

where for x ∈ [−R,R] the remainder term is

En(x) =f (n+1)(s)

(n+ 1)!xn+1

for some s between 0 and x. This implies that En(x) = O(xn+1), which isstronger than o(xn).

Note that a given function may have several asymptotic expansions, and thatan asymptotic series does not need to represent a specific function. However,the coefficients an in an asymptotic expansion of a function with respect to agiven asymptotic sequence are unique.

Example 2.5. Suppose that

F (λ) ∼∞∑n=0

anλn

as λ −→∞ and Reλ > 0.If the term e−λ is added to F (λ) the same asymptotic expansion is still valid,i.e.

e−λ + F (λ) ∼∞∑n=0

anλn

as λ −→∞, with Reλ > 0.We say that e−λ is ”beyond all orders” with respect to the asymptotic sequence{λ−n

}∞n=0

.

5

2.2 Watson’s lemma

There are many different techniques to obtain asymptotic expansions of inte-grals. One elementary method is integration by parts. In this section a partic-ular case of the integral (1.1) will be considered, namely

F (λ) =

∫ T

0

e−λtφ(t)dt. (2.1)

This corresponds to R(t) = −t, so R(t) has a maximum at the left endpoint ofthe interval [0, T ].

Example 2.6. (Integration by parts.) Suppose

F (λ) =

∫ T

0

e−λtg(t)dt,

where g ∈ C∞ and T ∈ R+. By using integration by parts it follows that theasymptotic expansion of this integral is

F (λ) ∼∞∑n=0

g(n)(0)

λn+1(2.2)

as λ −→ ∞. Indeed, making the first step with this well-known method oneobtains the following:

F (λ) = −g(t)e−λt

λ

∣∣∣T0

+

∫ T

0

g′(t)e−λt

λdt.

Continuing repeating this gives

F (λ) =

N∑n=0

(g(n)(0)

λn+1− g(n)(T )e−λT

λn+1

)+

∫ T

0

g(n+1)(t)e−λt

λn+1dt, (2.3)

where the integral in (2.3) is O(λ−(N+2)) as λ −→∞. This implies (2.2)

Clearly this method is not suited for all types of integrals of the form (1.1).Another way of finding the asymptotic expansion for integrals is the followingclassical result known as Watson’s lemma, that is a generalization of the exam-ple above.

Proposition 2.7. (Watson’s lemma.) Suppose that T ∈ R+ and that g(t) is acomplex-valued, absolutely integrable function on [0, T ]:∫ T

0

|g(t)|dt <∞.

Suppose further that g(t) is of the form g(t) = tσφ(t) where σ > −1 and φ(t)has an infinite number of derivatives in some neighborhood of t = 0. Then theexponential integral

F (λ) =

∫ T

0

e−λtg(t)dt

6

is finite for all λ > 0 and it has the asymptotic expansion

F (λ) ∼∞∑n=0

φ(n)(0)Γ(σ + n+ 1)

n!λ(σ+n+1)(2.4)

as λ −→ +∞.

For the proof of this proposition, see [8].

Remark 2.8. In many situations T = +∞. In this case the conditions areunnecessarily restrictive. If T = +∞ it suffices to assume that φ(t) = O(eat) ast→∞ for some a ∈ R, and the proposition still holds true.

Remark 2.9. From the definition of the gamma function

Γ(z) =

∫ ∞0

tz−1e−tdt, Re z > 0, (2.5)

it follows thatΓ(z + 1) = zΓ(z), (2.6)

and in particular Γ(n + 1) = n! for n ∈ N. So in case σ = 0 the asymptoticexpansion expression (2.4) is reduced to

F (λ) ∼∞∑n=0

g(n)(0)

λn+1,

which is the same result as (2.2).

Note that the integral expression (2.1) and the Laplace transform of f(t) definedby

L(f)(s) =

∫ ∞0

e−stf(t)dt,

are of the same form, with T = ∞. So in this case one can get an asymptoticexpansion of a Laplace transform for large s immediately. For many exponentialintegrals Watson’s lemma can be used by making adjustments of the variablesso that the required structure of the lemma is met, and then apply it to obtaina full asymptotic expansion.

Example 2.10. Consider an integral of the form

F (λ) =

∫ β

−αe−λt

2

φ(t)dt, (2.7)

where as in Proposition 2.7 the function φ is assumed to be infinitely differen-tiable in a neighborhood of 0. Let γ = min (α, β), then

F (λ)−∫ γ

−γe−λt

2

φ(t)dt = o(λ−n),

7

for all n ∈ N, as λ −→∞.Because removing the part of the interval that not is around the point t = 0 willgive an error beyond all orders with respect to

{λ−n

}∞n=0

. Now, using symmetry∫ γ

−γe−λt

2

φ(t)dt =

∫ γ

0

e−λt2

φ(t)dt+

∫ 0

−γe−λt

2

φ(t)dt

=

∫ γ

0

e−λt2

(φ(t) + φ(−t))dt = 2

∫ γ

0

e−λt2

φeven(t)dt, (2.8)

where

φeven(t) =(φ(t) + φ(−t))

2.

Making the substitution t =√s, the integral (2.8) becomes∫ γ2

0

e−λsφeven(

√s)√

sds. (2.9)

By writing φeven(t) as its Taylor series we obtain

φeven(t) ∼ 1

2

( ∞∑n=0

φ(n)(0)

n!tn +

φ(n)(0)

n!(−t)n

)

=1

2

∞∑n=0

φ(n)(0)

n!(1 + (−1)n)t2 =

∞∑n=0

φ(2n)(0)

(2n)!t2n,

which gives us

φeven(√s) ∼

∞∑n=0

φ(2n)(0)

(2n)!sn.

The integral (2.9) is now suitable for using Watson’s lemma and we observe thatσ = −1/2 and T = γ2 . Thus we obtain the asymptotic expansion

F (λ) ∼∞∑n=0

φ(2n)(0)Γ(n+ 1/2)

(2n)!λn+1/2(2.10)

as λ −→∞.This asymptotic expansion can be written in another form. Using the relation(2.6) for the Gamma function in (2.10), we get

Γ(n+ 1/2) =(2n)!

22nn!Γ(1/2).

From the definition of the Gamma function in (2.5), Γ(1/2) is written as aGaussian integral

Γ(1/2) =

∫ ∞0

t−1/2e−tdt =

∫ ∞−∞

e−s2

ds,

with t = s2. Integrating using polar coordinates, it follows that

Γ(1/2)2 =

(∫ ∞−∞

e−s2

ds

)2

8

=

∫ ∞−∞

∫ ∞−∞

e−(x2+y2)dxdy =

∫ π

−π

∫ ∞0

re−r2

drdθ

= 2π

∫ ∞0

re−r2

dr = π

∫ ∞0

e−ωdω = π.

So Γ(1/2) =√π. Thus the asymptotic expansion formula becomes

F (λ) ∼√π

λ

∞∑n=0

φ(2n)(0)

22nn!λ−n (2.11)

as λ −→∞.

This asymptotic expansion (2.11) illustrates the case when the exponent R(t)attains a maximum at an interior point of the interval of integration and as wewill see this result have resemblance to the leading order behavior obtained byLaplace’s method. This topic will be discussed in the next section.

9

3 Laplace’s Method

Consider an integral of the form

F (λ) =

∫ b

a

eλR(t)g(t)dt, (3.1)

where (a, b) is finite or infinite and real, as λ −→ +∞ . Laplace’s method isa way of finding approximations of integrals of this form. The idea is that themain contributions to F (λ) as λ −→ ∞ will come from a neighborhood of thepoint or points where R(t) has its maximum value on the interval (a, b), andthe rest of the contributions of the integral that are not close to the maximumwill be exponentially small in comparison.

Suppose first that R(t) has a maximum attained at a single interior point onthe interval (a, b), called tmax. Assume furthermore that R(t) and g(t) are in-finitely differentiable functions in a neighborhood of t = tmax and R′′(tmax) < 0.

Since we know that the main contribution of the integral (3.1) comes fromthe neighborhood of tmax, we can remove the rest of the interval because it’scontribution is beyond all orders with respect to

{λ−n

}∞n=0

. We write R(t) =R(t) − R(tmax) and choose δ as small as we please. Hence, the integral thatcontributes to the asymptotic expansion is

eλR(tmax)

∫ tmax+δ

tmax−δe−λR(t)g(t)dt

= eλR(tmax)

∫ δ

−δe−λR(tmax+τ)g(tmax + τ)dτ. (3.2)

For this integral we wish to make the change of variables,

R(tmax + τ) = −s2, (3.3)

in order to have (3.2) in the same form as the integral in Example 2.10 in theprevious section. First we need to solve (3.3) for τ as a smooth function of snear the point s = 0. Let

ψ(s, τ) = R(tmax + τ) + s2.

The implicit function theorem in this case fails because ∂ψ∂τ (0, 0) = 0. We go

around this problem by using a new variable v, defined by τ = sv. We thenhave

R(tmax + sv)

s2= −1. (3.4)

By expanding R(tmax + τ) in a neighborhood of τ = 0, we get

R(tmax + sv) =1

2R′′(tmax)(sv)2 +O((sv)3).

Using the expression above in the formula (3.4), one obtains

1

2R′′(tmax)v2 +O(sv3) = −1.

10

Observe that this equation is satisfied by the point (0, v0), where

v0 =

√−2

R′′(tmax). (3.5)

Note that we here choose v0 positive. Now, let

f(s, v) =

R(tmax + sv)

s2+ 1, if s 6= 0,

1

2R′′(tmax)v2 + 1, if s = 0.

(3.6)

Then f ∈ C∞. Since ∂f∂v (0, v0) = R′′(tmax)v0 6= 0, by the implicit function

theorem the relation f(s, v) = 0 locally near (0, v0) defines v as a function of s.Now introduce the change of variables τ(s) = sv(s). Then the integral in (3.2)becomes

eλR(tmax)

∫ β

−αe−λs

2

φ(s)ds,

where α =√−R(tmax − δ) and β =

√−R(tmax + δ). We also have φ(s) =

g(tmax + sv(s))(sv′(s) + v(s)). This is now an integral of the same form as (2.7)in the last section. By applying (2.11), we find that the asymptotic expansionis

F (λ) ∼ eλR(tmax)

√π

λ

∞∑n=0

φ(2n)(0)

22nn!λn(3.7)

as λ −→∞.Note that it is possible to calculate as many terms in the asymptotic expansion(3.7) as one likes, but the higher order derivatives of φ will quickly lead to longcalculations. For the first term of (3.7), we calculate φ(0):

φ(0) = g(tmax)v(0) =

√−2

R′′(tmax)g(tmax).

It follows that the leading order behavior of F (λ) is

F (λ) =

∫ b

a

e−λR(t)g(t)dt = eλR(tmax)

(√−2π

λR′′(tmax)g(tmax) +O(λ−3/2)

)(3.8)

as λ −→∞. See [8] for more terms of the expansion.

For an interval containing many interior points (t1, ..., tN ) at which the maxi-mum is attained, a similar result holds; one only has to sum the contributionsfrom each maximum point. Thus the leading order term in this case will begiven by

F (λ) = eλR(tmax)

N∑j=1

√−2π

λR′′(tj)g(tj) +O(λ−3/2)

as λ −→∞.

11

In case the maximum is obtained at one of the endpoints of the interval, the eval-uation of the leading order behavior is even simpler. By replacing R(t) with thefirst terms of its series expansion near the endpoint R(tmax) + R′(tmax)(t−tmax)will give the following results:

If the unique maximum is at the left boundary t = a of the interval [a, b] andR′(a) < 0, then

F (λ) = −eλR(tmax)

(g(a)

λR′(a)+O(λ−2)

)as λ −→∞.If the unique maximum is at the right boundary point t = b and R′(b) > 0, then

F (λ) = eλR(b)

(g(b)

λR′(b)+O(λ−2)

)as λ −→∞.

In the case when there are contributions coming from both interior and bound-ary points the dominant contribution will come from the interior points sincethe contribution from the boundary points is much smaller in comparison. WithLaplace’s method there is no limit in how many terms of the asymptotic expan-sion one can obtain, but the larger the number of terms there is the more effortit takes to calculate those terms. So it is in theory possible to obtain the fullasymptotic expansion of an integral using this method.

Throughout the rest of this paper only the leading order term of the asymptoticexpansion of integrals like (1.1) will be calculated since it gives an approxima-tion that for most applications is sufficient.

Example 3.1. Consider

n! = Γ(n+ 1) =

∫ ∞0

e−ttndt.

By making the variable substitution t = sn and rewriting sn as a exponentialfunction en(log(s)), the integral takes the form

Γ(n+ 1) = nn+1

∫ ∞0

en(log(s)−s)ds.

The integral is now of the right form needed to be able to apply Laplace’smethod. Set R(s) = log(s)− s. Since R′(s) = 1

s − 1 the maximum is at a pointsmax = 1 and since R′′(smax) = −1 the formula (3.8) can be used and thus theleading order behavior is

n! ∼ nn+1e−n

(√2π

n+O(n−3/2)

)as n −→∞.This result is the first term in Stirling’s formula that is used for approximatingfactorials.

12

4 Method of Steepest Descent

4.1 The idea

So far Laplace type integrals of real-valued functions have been studied, but formany problems and applications the integral is in the complex plane. Consideran integral of the form

F (λ) =

∫C

eλh(z)g(z)dz =

∫ b

a

eλh(z(t))g(z(t))z′(t)dt (4.1)

as λ −→ +∞.Here C is a smooth curve in the complex plane and z(t) = x(t)+ iy(t), t ∈ [a, b],a parametrization of the contour. We moreover assume that h(z) and g(z) arecomplex-valued entire functions of z.

Write h(z) = R(x, y) + iI(x, y) for z = x+ iy. Suppose that I(x, y) is constantalong C. Then the function F (λ) can be rewritten in the form

F (λ) = eiλI∫ b

a

eλR(x(t),y(t))u(t)dt+ ieiλI∫ b

a

eλR(x(t),y(t))v(t)dt, (4.2)

where u(t) = Re g(z(t))z′(t) and v(t) = Im g(z(t))z′(t). The asymptotic expan-sion of each of the integrals can here be evaluated using Laplace’s method sincethey are real. Note that it is highly unusual for an integral to have the propertythat I(x, y) is constant from the beginning, so in most cases it is not possibleto use Laplace’s method directly.

There is however a way of finding the asymptotic expansion of complex-valuedintegrals when I(x, y) not is constant along C. For a general contour C, thestrategy is to deform C into another new contour C ′ along which the imaginarypart I(x, y) of h(z) is constant.By Cauchy’s theorem it is possible to make contour deformations without af-fecting the value of the integral. When a suitable deformed contour C ′ hasbeen obtained an asymptotic expansion of the integral can be computed usingLaplace’s method. This method is called the method of steepest descent, or thesaddle point method for reasons to be explained.

Remark 4.1. Deformation of the contour is still feasible even if g(z) is merelymeromorphic, or even has branch cuts, by applying the residue theory.

4.2 Steepest descent paths and saddle points

Let z0 = x0 + iy0 be a point in the complex plane and suppose that h(z) =R(x, y) + iI(x, y) is analytic at z0. When the value of R(x, y) is declining fromits value in the point z0 in a direction emanating from z0, that direction is adescent direction.A curve emanating from z0 with the tangents being descent directions for every

13

point on the curve is a descent path. The descent paths with the most rapiddescent are called steepest descent paths. We desire to find these paths becauseit turns out that the curves where I(x, y) is constant at the same time are steep-est descent paths.

The gradient of a function indicates the direction in which the function increasesmost rapidly. As I(x, y) is constant on C ′, then the gradient of I(x, y):

∇I(x, y) =∂I

∂x(x, y)i +

∂I

∂y(x, y)j

is normal to C ′. Since h(z) is analytic the Cauchy-Riemann equations aresatisfied

∂R

∂x(x, y) =

∂I

∂y(x, y),

∂R

∂y(x, y) = −∂I

∂x(x, y).

This implies that

∂R

∂x(x, y)

∂I

∂x(x, y) +

∂R

∂y(x, y)

∂I

∂y(x, y) = 0, (4.3)

or, in other words, ∇R(x, y) · ∇I(x, y) = 0.

It follows that ∇R(x, y) is perpendicular to ∇I(x, y). Consequently ∇R(x, y)will be tangent to C ′. Thus, a properly oriented curve I(x, y) = I(x0, y0) rep-resents a curve of steepest descent of R(x, y). This explains the name ”Methodof steepest descent”.

We now explain why this method is also called the saddle point method. Supposet0 is a critical point of R(x(t), y(t)), that is

d

dtR(x(t), y(t))

∣∣∣∣t=t0

= 0,

or equivalently∇R(x(t), y(t)) · (x′(t), y′(t))|t=t0 = 0.

Since (x′(t0), y′(t0)) 6= 0, this implies that ∇R(x0, y0) = 0 where (x0, y0) =(x(t0), y(t0)). Using the Cauchy-Riemann equations we find that

∂R

∂x(x0, y0) = 0,

∂I

∂x(x0, y0) = 0

∂R

∂y(x0, y0) = 0,

∂I

∂y(x0, y0) = 0.

Hence both vectors ∇R(x, y) and ∇I(x, y) will vanish at this point;

∇R(x0, y0) = ∇I(x0, y0) = 0. (4.4)

So h′(z0) = 0. From the result (4.4) it follows that the main contribution comesfrom critical points of h(z). In particular, if R(x, y) has a maximum on C ′ at

14

z0, then z0 is a critical point of h.

Note that R(x, y) and I(x, y) are solutions to Laplace equation and thereforeare harmonic functions (which can be proved by Cauchy-Riemann’s equations).For harmonic functions the maximum principle states the following:

Theorem 4.2. (Weak Maximum Principle.) Let D be a bounded domain , andlet u(x, y) ∈ C2(D) ∩ C(D) be a harmonic function in D. Then the maximum(minimum) of u in D is achieved on the boundary ∂D.

Theorem 4.3. (Strong Maximum Principle.) Let u be a harmonic function inthe domain D. If u attains its maximum (minimum) at an interior point of D,then u is constant.

So R(x, y) cannot have a maximum at (x0, y0), i.e. the critical point (x0, y0) ofR(x, y) is a saddle point. Because of the importance of the saddle points in theevaluation of integrals with contours without endpoints, the method of steepestdescent is also called the saddle point method.

Near the saddle point z = z0 the Taylor expansion of h(z) has the form

h(z) = h(z0) +h(p)(z0)

p!(z − z0)p +O((z − z0)p+1). (4.5)

By setting z = z0 + reiθ and h(p)(z0) = aeiα, the leading terms of the Taylorexpansion of h(z) in (4.5) may be written in the form

h(z)− h(z0) =aeiα

p!rpeipθ +O(rp+1) =

=rpa

p!(cos(α+ pθ) + i sin(α+ pθ)) +O(rp+1).

This gives the following when dividing it into real and imaginary parts

R(x, y) =rpa

p!cos(α+ pθ) +O(rp+1)

and

I(x, y) =rpa

p!sin(α+ pθ) +O(rp+1).

As discussed before it is known that the steepest descent curves that go throughthe saddle point have the property that Im (h(z)− h(z0)) = 0, therefore

sin(α+ pθ) = 0.

This implies that the direction of the steepest descent curves comes from thefollowing angles:

θ = −αp

+nπ

p, (4.6)

15

where n is any odd number n = 1, 3, 5, ..., 2p− 1. The steepest ascent directionsare obtained from the even numbers n = 0, 2, ..., 2p− 2. We will only be inter-ested in the case of simple saddle points, i.e. p = 2.

Example 4.4. Let h(z) = i sin(z). Then since h′(z) = i cos(z) and the saddlepoints are

zm =π

2+mπ, m ∈ Z.

Consider z0 = π2 . Then h′′(z0) = −i and thus we obtain α = −π2 . So by (4.6)

the steepest descent directions are

θ = −π4

+nπ

2, n = 1, 3.

4.3 Leading order behavior of F (λ)

Recall the expression (4.2) where the Laplace’s method could be used becausethe imaginary part of h(z) was set to be constant on C. Now suppose that C isdeformed to go where the imaginary part is constant along a steepest descentpath and through a saddle point as described earlier. Because the imaginarypart is constant, h(z)− h(z0) is real on this deformed contour C ′.

Let the maximum value of the real-part of h(z) appear in a point denoted t0on the interval [a, b] and that the saddle point z0 = z(t0) is simple (that ish′(z0) = 0 and h′′(z0) 6= 0). Furthermore let C ′ have the parametrization z(t),making it resemble the interval [a, b]. The integral (4.1) thus becomes

F (λ) = eλh(z0)∫ b

a

eλ(h(z(t))−(h(z0)))g(z(t))z′(t)dt.

From the leading order behavior expression (3.8) when the maximum valuecomes from an interior point, we have that∫ b

a

eλ(h(z)−(h(z0)))g(z(t))z′(t)dt =

=

√−2π

λ(z′(t0))2h′′(z0)g(z0)z′(t0) +O(λ−3/2). (4.7)

Becaused2

dt2(h(z)− h(z0)) = (z′(t))2h′′(z(t)) + h′(z(t))z′′(t),

it follows thatd2

dt2(h(z)− h(z0)) = (z′(t))2h′′(z(t)), (4.8)

since the second term is zero because h′(z0) = 0. Note that (4.8) is a negativereal number. We rewrite√

−2π

(z′(t0))2h′′(z0)z′(t0) =

√2π

|h′′(z0)|

16

and with z′(t0) = |z′(t0)|eiθ(z0) we thus obtain the following formula for theasymptotic leading order behavior of F (λ):

F (λ) = λ−1/2eλh(z0)

(eiθ(z0)

√2π

|h′′(z0)|g(z0) +O(λ−1)

)(4.9)

as λ −→∞.

17

5 Applications

5.1 Airy functions

These are special functions named after George Biddell Airy that are solutionsto the differential equation

y′′(x)− xy(x) = 0, (5.1)

called the Airy equation. The Airy functions appear in many situations in bothmathematics and physics1. For example, in quantum mechanics, they describethe behavior of the wave function near the classical turning points. To later beable to examine the asymptotic behavior of the Airy function, suppose that thesolutions to the differential equation (5.1) can be written as contour integralsof the form

y(x) =1

2πi

∫C

q(z)exzdz. (5.2)

Substituting the integral (5.2) into Airy’s equation (5.1) we obtain∫C

z2q(z)exzdz −∫C

xq(z)exzdz = 0. (5.3)

Then, by using integration by parts, we get∫C

z2q(z)exzdz − q(z)exz∣∣∣∣C

+

∫C

q′(z)exz = 0.

This is done because we don’t want the parameter x as a factor in the secondintegral of (5.3). Let’s choose q(z) so that it solves the differential equation

q′(z) + z2q(z) = 0,

i.e. take q(z) = e−z3/3. We also need to choose the contour C in such a way

that the integrand disappears at the end-points of C. Therefore C is set to goto infinity in a region where Re (z3) = r3 cos(3θ) > 0 (with z = reiθ). This givesthree sectors defined by

−5π

6< θ <

−π2,

−π6

< θ <π

6,

π

2< θ <

5π

6.

In each of these sectors the integrand goes to zero as z goes to infinity. Theintegrand of (5.2) thus is exz−z

3/3 and we have a solution to the differentialequation (5.1) of the form

y(x) =1

2πi

∫C

exz−z3

3 dz. (5.4)

1Airy functions and applications to physics, Olivier Valle’e Manuel Soares.

18

Figure 1: The sectors − 5π6 < θ < −π2 ,−

π6 < θ < π

6 ,π2 < θ < 5π

6 are shown ingrey and the contours C1, C2, and C3 in blue color.

To get non-trivial solutions, we let C go from infinity in one sector and end atinfinity in another. Because there are three sectors we have essentially threedifferent choices of contours, C1, C2, and C3. Each choice gives a solution yi.Since (5.1) only has two linearly independent solutions these are of course lin-early dependent; clearly y1 + y2 + y3 = 0. See Figure 1.

The solution y1 is called the Airy function of the first kind and is denoted by

Ai (x) =1

2πi

∫C1

exz−z3/3dz. (5.5)

The Airy function of the second kind, also known as the Bairy function, isdefined by

Bi (x) = iy2 − iy3,

or in other words is the integral (5.4) over C2−C3. These functions are linearlyindependent solutions to the differential equation (5.1).

19

5.1.1 Asymptotics as x −→ +∞

Let us write the parameter x in polar form as x = reik. We here consider theasymptotic behavior of Ai (reik) as r −→ ∞ for a fixed value of k ∈ (−π, π).This includes the case x −→ +∞. Since the integral (5.5) is not of the form (4.1)we first introduce the change of variables z =

√ru, which gives the integrand

the appropriate form:

Ai (reik) =

√r

2πi

∫C1

er3/2(eiku−u3/3)du (5.6)

where r −→∞.To simplify the calculations later we set λ = r3/2 and consider the integral

I(λ) =

∫C1

eλ(eiku−u33 )du.

Observe that this is precisely of the form (4.1). We here have that

h(u) = eiku− u3

3. (5.7)

We know that saddle points are obtained from h′(u) = eik−u2 = 0, which givestwo saddle points at u = ±eik/2. We will be using the notation

uR = eik/2, uL = −eik/2

for the two saddle points since the former is located in the right halfplane andthe latter in the left. Now, writing h(uR) = 2

3e3ik/2 in terms of cosine and sine

we get

Re (h(uR)) =2

3cos(

3k

2), Im (h(uR)) =

2

3sin(

3k

2). (5.8)

In the same way we get

Re (h(uL)) = −2

3cos(

3k

2), Im (h(uL)) = −2

3sin(

3k

2). (5.9)

The formulas (5.8) and (5.9) will be helpful in determining from where thedominant contribution to the asymptotic expansion will come. We observe that,

Re (h(uL)) < Re (h(uR)) for − π

3< k <

π

3,

Re (h(uR)) < Re (h(uL)) for − π < k < −π3,

π

3< k < π,

andRe (h(uR)) = Re (h(uL)) for k = π.

For the imaginary part we have

Im (h(uR)) = Im (h(uL)) for k = 0, k = −2π

3, k =

2π

3.

20

To get a better view of how the steepest descent paths and the level curvesof R(x, y) depend on the parameter k, we start by considering some particularvalues of k:

Consider first k = 0. For this value of k we deform C1 to the steepest descentpath that goes through uL, which is part of the contour I(x, y) = Im (h(uL)) =0. Note that it is not possible to deform the contour C1 to the steepest descentpath through the saddle point uR. This value of k is interesting because it isone of the few cases where the steepest descent curves can be drawed by hand.Examining h(u) in (5.7) with k = 0, we find that

Im (h(u)) = b

(b2

3− a2 + 1

), u = a+ ib.

Clearly, the zero set of this expression is the union of the hyperbola and line inFigure 2.

Figure 2: For k = 0. Left: Contour plot for Re (h(u)), darker color indicatelower level curves. Right: The steepest descent paths.

For k = π2 the contour C1 is still suitable for deformation and the dominant

contribution is coming from uL, see Figure 3.

21

Figure 3: For k = π2 . Left: Contour plot for Re (h(u)), darker color indicate

lower level curves. Right: The steepest descent paths.

When k = 2π3 the steepest descent curve look as in Figure 4. In this case the

steepest descent contour will go through uL to uR and then turn left to go alongthe descent path down from this saddle point. The real part of uL is greaterthan the real part of uR, so the dominant contribution comes from uL.

Figure 4: For k = 2π3 . Left: Contour plot for Re (h(u)), darker color indicate

lower level curves. Right: The steepest descent paths.

Back to finding the leading order formula. We have

h′′(u) = −2u so h′′(uL) = 2eik/2,

which gives α = k2 . By the formula (4.6) the steepest descent directions are

θ = −k4

+nπ

2,

22

for odd n. Formula (4.9) gives that the leading order behavior of I(λ) is

I(λ) = λ−1/2eλh(uL)

(eiθ(uL)

√2π

|h′′(uL)|g(uL) +O(λ−1)

)

=√πλ−1/2e−

2λe3ik/2

3 ei(−k4+

π2 )(1 +O(λ−1)) (5.10)

as λ −→∞.Hence by setting λ = r3/2 and using the leading order expression of I(λ) fromabove in the integral (5.6), the Airy function has the asymptotics

Ai (x) =1

2x1/4√πe−

2x3/2

3 (1 +O(|x|− 32 )) (5.11)

as x −→∞.Note that x ∈ C and that both x1/4 and x3/2 are the principal branches in(5.11). In a similar way the leading order expression for the Bairy function isobtained to be

Bi (x) =1

x1/4√πe

2x3/2

3 (1 +O(|x|− 32 )) (5.12)

as x −→∞.

23

5.1.2 Asymptotics as x −→ −∞

When k = π we have steepest descent paths with the appearance as in Figure 5.We here have two contours to consider. The first is going through uL ending inthe right sector and the second is beginning in the right sector going through uR.The contributions from both of these saddle points must be taken into accountwhen evaluating the asymptotics. This is because the contributions from bothsaddle points are of the same order. Note that the leading order asymptotics isoscillatory, since h(uL) and h(uR) are both purely imaginary.

Figure 5: For k = π. Left: Contour plot for Re (h(u)), darker color indicatelower level curves. Right: The steepest descent paths.

For this value of k we choose a contour C ′1 containing both the steepest descentcurve through uL and the steepest descent curve through uR. We write

C ′1 = γL + γR.

By (5.6) we have

Ai (reiπ) =

√r

2πi

∫C′

1

er3/2(eiπu−u3/3)du

=

√r

2πi

(∫γL

er3/2(eiπu−u3/3)du+

∫γR

er3/2(eiπu−u3/3)du

)= Ai uL(reiπ) + Ai uR(reiπ). (5.13)

From the formula (5.10) with k = π the leading order behavior of IuL(λ) is

IuL(λ) =√πλ−1/2e−

2λe3iπ/2

3 ei(−π4 +π

2 )(1 +O(λ−1)) (5.14)

as λ −→∞.Since λ = r3/2, the first term in (5.13) has the leading order asymptotics

Ai uL(x) =1

2i|x|1/4√πei

(2|x|3/2

3 +π4

)(1 +O(|x|− 3

2 ) (5.15)

24

as x −→ −∞.

Becauseh′′(u) = −2u and h′′(uR) = −2eik/2,

we obtain α = −k2 for this saddle point. By the formula (4.6) the steepestdescent directions are

θ =π

4+nπ

2.

By the formula (5.10) the leading order behavior of IuR(λ) becomes

IuR(λ) =√πλ−1/2e

2λe3iπ/2

3 ei(−π4 +π)(1 +O(λ−1)) (5.16)

as λ −→∞.Again using that λ = r3/2, we find that the second term in (5.13) has the leadingorder asymptotics

Ai uR(x) = − 1

2i|x|1/4√πe−i

(2|x|3/2

3 +π4

)(1 +O(|x|− 3

2 )) (5.17)

as x −→ −∞.

From (5.15) and (5.17) we obtain the asymptotic expansion of Ai (x):

Ai (reiπ) = Ai uL(reiπ) + Ai uR(reiπ)

=1

2i|x|1/4√π

(ei

(2|x|3/2

3 +π4

)− e−i

(2|x|3/2

3 +π4

))(1 +O(|x|− 3

2 )).

Thus we finally get

Ai (x) =1

|x|1/4√π

sin

(2|x|3/2

3+π

4

)(1 +O(|x|− 3

2 )) (5.18)

as x −→ −∞.Similarly for the Bairy function, we obtain

Bi (x) =1

|x|1/4√π

cos

(2|x|3/2

3+π

4

)(1 +O(|x|− 3

2 )) (5.19)

as x −→ −∞.

Note the drastic difference in the leading order behavior of the Airy and Bairyfunctions when k = π compared to the previous case. The asymptotic behav-ior of a function in the complex plane may differ between particular sectorseven though the function considered is entire. This is known as the Stokes phe-nomenon. In our case this happens precisely along the negative real line. Formore detail about the Stokes phenomenon, see [6].

Figure 6 below shows a plot of the Airy and Bairy functions.

25

Figure 6: The Airy function (solid line) and the Bairy function (dashed line).

5.2 The KdV equation

The Korteweg–de Vries equation,

ut + uxxx + 6uux = 0, (5.20)

named after Diederik Korteweg and Gustav de Vries is describing the behaviorof waves in shallow water. The behavior of solutions of the KdV equation isstill an area of active research. We will here examine the Cauchy problem forthe linearized KdV equation,

ut + uxxx = 0, (5.21)

with the initial data u(x, 0) = g(x). We assume that g belongs to C∞0 (R), thatis the class of infinitely differentiable functions with compact support, whichimplies that g can be extended to an entire function. One can solve this equationusing the Fourier transform. Indeed, taking the Fourier transform with respectto x we obtain

ut + (iω)3u = 0 and u(ω, 0) = g(ω).

This implies that

u(ω, t) = g(ω)e−(iω)3t.

26

We thus obtain the solution

u(x, t) =1

2π

∫ ∞−∞

g(ω)eiω3t+iωxdω. (5.22)

Writing

h(ω) = i(ω3 +

ωx

t

),

the integral (5.22) becomes

u(x, t) =1

2π

∫ ∞−∞

g(ω)eth(ω)dω. (5.23)

Note that this integral now is of the same form as (4.1). We wish to find thelong-time behavior of u(x, t) as t −→∞ with x

t fixed. We first handle the casesxt < 0 and x

t > 0.

We begin with the case xt < 0. Examining the exponent function h in integral

(5.23), we find the saddle points

ωL = −∣∣∣ x3t

∣∣∣1/2 and ωR =∣∣∣ x3t

∣∣∣1/2 .In this case we deform the real line to the union of two contours, see Figure 7.The first begins in the valley of the upper left half plane, goes through ωL andends close to the imaginary axis in the valley in the lower half plane. The secondcontour begins in the valley in the lower half plane, goes through ωR and endsin the valley in the first quadrant. The contributions from the saddle points areof the same order, therefore both these contributions must be taken into account.

Figure 7: Left: Contour plot for Re (h(u)), darker color indicate lower levelcurves. Right: The steepest descent paths.

We writeu(x, t) = uωL(x, t) + uωR(x, t). (5.24)

27

Because h′′(ω) = 6iω, then

h′′(ωL) = −6i∣∣∣ x3t

∣∣∣1/2 and h′′(ωR) = 6i∣∣∣ x3t

∣∣∣1/2 (5.25)

We begin with evaluating the contribution coming from ωL. From (5.25) wefind that α = 3π

2 , thus by the formula (4.6) the steepest descent directions are

θ = −3π

4+nπ

2.

From the formula (4.9) the leading order behavior of uωL(x, t) in (5.24) is ob-tained:

1√

12πt∣∣ x3t

∣∣1/4 g(−∣∣∣ x3t

∣∣∣1/2) e2it| x3t |3/2−iπ4 (1 +O(t−1)) (5.26)

as t −→∞.For ωR we get α = π

2 by using (5.25). From the formula (4.6) the steepestdescent directions are

θ = −π4

+nπ

2.

From the formula (4.9) we obtain the leading order behavior of uωR(x, t) in(5.24):

1√

12πt∣∣ x3t

∣∣1/4 g(∣∣∣ x

3t

∣∣∣1/2) e−2it| x3t |3/2+iπ4 (1 +O(t−1)) (5.27)

as t −→∞.Using the complex conjugate of the fourier transform g(−ω) = g(ω) and writing

g(∣∣ x3t

∣∣1/2) = r(∣∣xt

∣∣)eik( xt ), then by adding these formulas we obtain the leadingorder behavior of u(x, t):

u(x, t) =r(|xt |)√πt∣∣ 3xt

∣∣1/4 cos

(2t∣∣∣ x3t

∣∣∣3/2 − π

4− k

(xt

))(1 +O(t−1)) (5.28)

as t −→∞.

For xt > 0 we have the saddle points

ωL = −i( x

3t

)1/2and ωR = i

( x3t

)1/2.

In this case the contour is deformed so it goes through ωR and end in the valleyin the first quadrant, see Figure 8. So the dominant contribution comes from ωR.

28

Figure 8: Left: Contour plot for Re (h(u)), darker color indicate lower levelcurves. Right: The steepest descent paths.

Because

h′′(ωR) = −6( x

3t

)1/2,

we have α = π and the steepest descent directions

θ = −π2

+nπ

2.

The leading order behavior of u(x, t) when xt > 0 thus is

u(x, t) =1√

12πt( x3t )1/4

g

(i( x

3t

)1/2)e−2t(

x3t )

3/2

(1 +O(t−1)) (5.29)

as t −→∞.Note that the asymptotics in the two cases considered are drastically different.The error terms depend on the parameter η = x

t .

For xt −→ 0, we begin by making the change of variables

−iz = ω(3t)1/3 (5.30)

in the integral (5.23) and write ν = x(3t)−1/3. The integral then becomes

u(x, t) =1

2πi(3t)1/3

∫ i∞

−i∞g

(−iz

(3t)1/3

)ezν−

z3

3 dz

=1

2πi(3t)1/3

∫C1

g

(−iz

(3t)1/3

)ezν−

z3

3 dz. (5.31)

Note the similarity between this integral and the integral (5.5). The leadingorder behavior of u(x, t) can be calculated by expanding g in a neighborhoodof z = 0 and using the leading order behavior of the Airy function. We thus

29

obtain the leading order behavior:

u(x, t) = (3t)−1/3g(0)Ai (ν)(1 +O((3t)−1/3)). (5.32)

as t −→∞.We see that it connects the asymptotics in (5.28) and (5.29). Given the structureof the prefactor in equation (5.32), it is worth mentioning that there is a relationbetween the Airy equation and the linearized KdV equation. Indeed, if we let

u(x, t) =1

(3t)1/3y

(x

(3t)1/3

), (5.33)

then a simple calculation shows that u satisfies (5.23) if y satisfies (5.1). Wetherefore expect a solution that depends on the scale x(3t)−1/3 as x

t −→ 0. Theequation (5.33) is an example of an ODE reduction of a PDE.

5.3 Concluding remarks

In the last sections of this paper we have discussed the method of steepest de-scent and some applications to linear ODE’s and linear PDE’s. The classicalsteepest descent method also has applications to other areas of mathematics,such as combinatorics. Indeed, the famous Hardy-Ramanujan-Uspensky for-mula for the partition function can be obtained by applying it, see [3]. In recentyears a nonlinear method of steepest descent for oscillatory Riemann-Hilbertproblems has been introduced by Deift and Zhou [7]. This method can be usedto compute the long-time behavior of solutions of so-called integrable nonlinearPDE’s, such as the KdV equation (5.20) and the nonlinear Schrodinger equa-tion. In fact, the nonlinear method of steepest descent was introduced with thepurpose of getting a better understanding of the nonlinear KdV equation. Thestudy of the behavior of integrable nonlinear PDEs continues to be an activearea of research.In recent years so-called Painleve transcendents has started to appear in manyapplications. For example, ODE reductions of integrable nonlinear PDE’s leadto Painleve equations. The Painleve transcendents are nonlinear analogs of theclassical special functions such as the Airy and Bessel functions. Also thesetranscendents can be studied with the nonlinear steepest descent method. In-deed, the nonlinear method of steepest descent has also been used to solvelong-standing problems in e.g. combinatorics, such as Ulam’s problem for thelongest increasing subsequence in random permutations, see [4].

6 Acknowledgments

I would like to thank my supervisor, Jorgen Ostensson, for his help and guidanceduring this project.

30

7 References

[1] Complex Variables: Introduction and Applications, Mark J. Ablowitz, A.S. Fokas, Cambridge University Press, 2003

[2] Handbook of mathematical functions, M. Abramowitz and I. Stegun, DoverPublications, New York, 1992. Reprint of the 1972 edition.

[3] Advanced Complex Analysis: A Comprehensive Course in Analysis, Part2B, Barry Simon, American Mathematical Soc., 2015

[4] Combinatorics and Random Matrix Theory, Jinho Baik, Percy Deift, TouficSuidan, American Mathematical Soc., 2016

[5] Asymptotic expansion of integrals, Norman Bleistein, RichardA.Handelsman, Ardent Media, 1975

[6] Advanced Mathematical Methods for Scientists and Engineers I: Asymp-totic Methods and Perturbation Theory, Carl M. Bender, Steven A. Orszag,Springer Science and Business Media, 1999

[7] A steepest descent method for oscillatory Riemann-Hilbert problems,P.Deift, X.Zhou, Ann. of Math. (2) 137, 1993

[8] Applied Asymptotic Analysis, Peter David Miller, American MathematicalSoc., 2006

[9] Airy Functions and Applications to Physics, Olivier Vallee, Manuel Soares,Imperial College Press, 2004

[10] Asymptotic Methods for Integrals, Nico M Temme, World Scientific, 2014

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